Problem: Compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$
Solution: We see that
\[\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix} = 2 \begin{pmatrix} \sqrt{3}/2 & -1/2 \\ 1/2 & \sqrt{3}/2 \end{pmatrix} = 2 \begin{pmatrix} \cos \frac{\pi}{6} & -\sin \frac{\pi}{6} \\ \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \end{pmatrix}.\]Note that $\begin{pmatrix} \cos \frac{\pi}{6} & -\sin \frac{\pi}{6} \\ \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \end{pmatrix}$ corresponds to a rotation of $\frac{\pi}{6}$ around the origin.

In general, for a rotation matrix,
$$
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}^k = \begin{pmatrix} \cos k\theta & -\sin k\theta \\
\sin k\theta & \cos k\theta
\end{pmatrix}.
$$Hence,
$$
\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6 = 2^6 \begin{pmatrix} \cos \frac{\pi}{6} & -\sin \frac{\pi}{6} \\ \sin \frac{\pi}{6} & \cos \frac{\pi}{6} \end{pmatrix}^6 = 2^6 \begin{pmatrix} \cos {\pi} & -\sin {\pi} \\ \sin {\pi} & \cos {\pi}\end{pmatrix} = \boxed{ \begin{pmatrix} -64 & 0 \\ 0 & -64 \end{pmatrix} }.
$$